Question

$P$ and $Q$ are the points of trisection of the diagonal $BD$ of the parallelogram $ABCD$, Prove that $CQ$ is parallel to $AP$. Prove also that $AC$ bisects $PQ$.

Answer 1

Given : In ||gm, $ABCD,P$ and $Q$ are the points of trisection of the diagonal $BD$.

To prove : (i) $CQ||AP$ and also $AC$ bisects $PQ$

Proof : Since, diagonals of a parallelogram bisect each other

$\therefore$ $AO=OC$ and $BO=OD$

$\therefore$ $P$ and $Q$ are point of trisection of $BD$

$\therefore$ $BP=PQ=QD$ ...(i)

$\because$ $BO=OD$ and $BP=QD$ ....(ii)
Subtracting , (ii) from (i) we get

$OB-BP=OD-QD$

$\Rightarrow$ $OP=OQ$

In quadrilateral $APCQ$,

$OA=OC$ and $OP=AQ$ (proved)

Diagonals $AC$ and $PQ$ bisect each other at $O$

$\therefore$ $APCQ$ is a parallelogram

Hence, $AP||CQ$.